1. The problem asks to find the inequality represented by the graph of an upward-opening parabola with vertex at about $(1, -4)$, crossing the x-axis near $x = -1$ and $x = 3$, and the shaded region is outside the parabola. 2. The general form of a parabola is given by the quadratic function: $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. 3. From the graph, the vertex is approximately $(1, -4)$, so the equation becomes: $$y = a(x - 1)^2 - 4$$ 4. The parabola crosses the x-axis near $x = -1$ and $x = 3$, so these are the roots. The roots satisfy: $$a(x - 1)^2 - 4 = 0$$ 5. Substitute $x = -1$: $$a(-1 - 1)^2 - 4 = 0 \Rightarrow a( -2)^2 - 4 = 0 \Rightarrow 4a - 4 = 0$$ 6. Solve for $a$: $$4a = 4 \Rightarrow a = 1$$ 7. So the parabola equation is: $$y = (x - 1)^2 - 4$$ 8. The parabola opens upward since $a = 1 > 0$. 9. The shaded region is outside the parabola, meaning the inequality is: $$y \leq (x - 1)^2 - 4$$ 10. However, since the shading is outside the parabola, the inequality is: $$y \geq (x - 1)^2 - 4$$ 11. Therefore, the inequality represented by the graph is: $$y \geq (x - 1)^2 - 4$$